Calculating the Rotational Inertia

1. The problem statement, all variables and given/known data
A rigid structure consisting of a circular hoop of radius R and mass m, and a square made of four thin bars, each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of 4.0 s. Assume R = 0.90 m and m = 3.0 kg, calculate the structure's rotational inertia about the axis of rotation.

3. The attempt at a solution
The third equation is constructed by noticing that the problem states that the structure consists of both the square made of thin rods and the hoop. The square only has two rods that have moments of inertia because two are perpendicular to the axis and the other two are parallel, which have a moment of inertia of 0 (at least that is what I have been told). So...

Ok, I don't why I put a 2*0.90 in the hoop's moment of inertia equation. That was a mistake. As for the horizontal rods, I have no idea and I need some explanation of them and what to do for them because I thought they had no moment of inertia. The rod furthest to the left must be then I = (1/12)(3 kg)(0.09 m)^2, but why not include its distance from the axis?

Staff: Mentor

As for the horizontal rods, I have no idea and I need some explanation of them and what to do for them because I thought they had no moment of inertia.

Realize that the rods are being rotated about a vertical axis. The equation you have, I_rods = (1/12)ML^2, can be used for a horizontal rod rotating about a vertical axis (through its center of mass). (Use the parallel axis theorem to move the axis.)

The rod furthest to the left must be then I = (1/12)(3 kg)(0.09 m)^2, but why not include its distance from the axis?

Just the opposite: you must include the distance from the axis. Trick question: What's the rotational inertia of a rod about an axis parallel to the rod (and through its center of mass)? Hint: Only three of the rods contribute to the total rotational inertia.

Staff: Mentor

Good. This applies to the left vertical rod. (What about the right one?)

I_hoop = (1/2)(3 kg)(0.09 m)^2 + (3 kg)(0.09 m)^2

Good.

I_horizontalRod = (1/12)(3 kg)(0.09 m)^2 - (3 kg)(0.09 m)^2

The horizontal rods have a moment of inertia of (1/12)(3 kg)(0.09 m)^2, but it is shifted the left of the axis by the length of a rod.

Careful. (1/12)(3 kg)(0.09 m)^2 is the rotational inertia for a horizontal rod rotating about a vertical axis through its center. What does the parallel axis theorem tell you to add to that to get its rotational inertia about the required axis? (Left and right don't matter--all that matters is the distance from cm to the axis.)